Question

Please solve the following:
1. y''=1+(y')^2
2. xdy-(3y+x^5*y^1/3)dx=0

Answer 1

Answer on Question #47686 – Math – Differential Calculus | Equations

Please solve the following:

1. $y'' = 1 + (y')^2$

2. $xdy - (3y + x^5 * y^1/3)dx = 0$

Solution.

1. $y'' = 1 + (y')^2$

$$\begin{array}{l} v(x) = y' \rightarrow v' = 1 + v^2 \rightarrow \frac{dv}{1 + v^2} = dx \rightarrow \arctan(v) = x + c_1 \\ \rightarrow \arctan(y') = x + c_1 \rightarrow \\ \end{array}$$ $$y' = \tan(x + c_1) \rightarrow y = \int \tan(x + c_1)dx = -\ln[\cos(x + c_1)] + c_2.$$ $$y = -\ln[\cos(x + c_1)] + c_2.$$

2. $xdy - (3y + x^5 y^1/3)dx = 0$

Verify that $y = 0$ is the solution to equation. Let $y \neq 0$ and

$$\begin{array}{l} xy' - 3y = x^5 y^1/3 \rightarrow \frac{y'}{y^1/3} - \frac{3}{x} y^2/3 = x^4 \rightarrow \left(y^2/3\right)' = \frac{2}{3y^1/3} y' \rightarrow \\ \rightarrow \frac{2y'}{3y^1/3} - \frac{2}{x} y^2/3 = \frac{2}{3} x^4 \rightarrow \\ \end{array}$$ $$\begin{array}{l} v(x) = y^2/3 \rightarrow v' - \frac{2v}{x} = \frac{2x^4}{3} \rightarrow \frac{v'}{x^2} - \frac{2v}{x^3} = \frac{2x^2}{3} \rightarrow \frac{d}{dx} \left(\frac{v}{x^2}\right) = \frac{2x^2}{3} \rightarrow \\ \rightarrow \frac{v}{x^2} = \frac{2x^3}{9} + c \rightarrow y^2/3 = \frac{2x^5}{9} + cx^2 \rightarrow y = \frac{x^3}{27} (2x^3 + 9c)^3/2 \\ y = \frac{x^3}{27} (2x^3 + 9c)^3/2 \text{ and } y = 0. \end{array}$$

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